Use the GCF of the terms to write the expression as the product of two factors with integer coefficients. -2x3 - 4x2 + 4x
What factor goes into all of these numbers?
2
That's one of them. What else?
1
x also goes into all of them
So the GCF is 2x
alright, so I need to make to seprate eqations with theb product of 2x?
Divide all of the numbers by your gcf
-2x3 - 4x2 + 4x -1x1-2x1+2x
You got the coefficients right, -2x^3/2x = -1x^2
x(-2x2 + 2x - 2)
Also pull out the factor of two
A. x(-2x2 - 4x + 4) B. -2x(x2 + 2x - 2) C. x(-2x2 + 2x - 2) D. -2x(-2x2 + 2x - 2)
If x(-2x^2 - 4x + 4) 2x(?)
If the problem were 3x^3 + 6x^2 + 9x the GCF is 3x so we will factor out the GCF from all of the numbers (3x)*x^2 + (3x)*2x + (3x)*3 = (3x)(x^2 + 2x + 3)
your so patient. I guess I just don't undestand. Ill read through again.
you can't devide three by two without leftovers.
Where did I divide three by two?
you didn't. NM
Well when we divide variables like x or x^2 we subtract the bottom power from the top one \[\Large \frac{x^n}{x^m}= x^{n-m}\] so if we divided x^3 by x we would get \[\Large \frac{x^3}{x^1}=x^{3-1}=x^2\] The distributive property says that if we have a number a, b, and c that \[a(b+c)=ab+ac\] So when we have something like 3x^3 + 6x^2 + 9x we divide each of the numbers by the gcf and put it on the outside kind of like with ab + ac our gcf is 'a' so we put it on the outside a(b + c) So \[\Large \frac{3x^3}{3x} = x^2\] \[\Large \frac{6x^2}{3x} = 2x\] \[\Large \frac{9x}{3x} =3\] When we put all of the numbers back after we have divided them by our gcf we get \[\Large x^2 + 2x + 3\] but we have to put the gcf on the outside so we end up with \[\Large 3x(x^2 + 2x + 3)\]
So my answer sounld be \[2x(x^{2} + 2x - 2)\]
Yes, but it seems they used -2x as their gcf, put a negative in front of your 2x and you will have the correct answer
k (wipes sweat from brow) thanks lol
lol :D No problem Good job
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